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Chapter 16: Problem 52

Phenylacetic acid $\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{COOH}\right)$ is one of the substances that accumulates in the blood of people with phenylketonuria, an inherited disorder that can cause mental retardation or even death. A \(0.085 \mathrm{M}\) solution of $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{COOH}$ has a pH of 2.68. Calculate the \(K_{a}\) value for this acid.

Short Answer

Expert verified
The $K_{a}$ value for phenylacetic acid is approximately \(5.28 \times 10^{-5}\).

Step by step solution

01

Calculate the concentration of H₃O⁺ ions

Using the formula for pH, we can calculate the concentration of H₃O⁺ ions in the solution: pH = -log[H₃O⁺] 2.68 = -log[H₃O⁺] First, we need to isolate the concentration of H₃O⁺ ions: [H₃O⁺] = 10^(-pH) [H₃O⁺] = 10^(-2.68) Now, we can calculate the concentration of H₃O⁺ ions: [H₃O⁺] ≈ 2.09 × 10⁻³ M.
02

Find the concentration of the conjugate base

Since phenylacetic acid (C₆H₅CH₂COOH) is a weak acid, it partially ionizes in water as follows: C₆H₅CH₂COOH (aq) + H₂O (l) ⇌ C₆H₅CH₂COO⁻ (aq) + H₃O⁺ (aq) Initially, we know the concentration of C₆H₅CH₂COOH is 0.085 M. After the reaction, the concentration of H₃O⁺ is 2.09 × 10⁻³ M, and this is also the concentration of the conjugate base, C₆H₅CH₂COO⁻.
03

Calculate the change in concentration of C₆H₅CH₂COOH

As the reaction progresses, some of the C₆H₅CH₂COOH molecules ionize, and their concentration decreases. Since we know the concentration of H₃O⁺ ions (2.09 × 10⁻³ M), we can determine the change in concentration of C₆H₅CH₂COOH: ΔC₆H₅CH₂COOH = -ΔH₃O⁺ ΔC₆H₅CH₂COOH ≈ -2.09 × 10⁻³ M At equilibrium, the concentration of C₆H₅CH₂COOH is 0.085 M - 2.09 × 10⁻³ M ≈ 0.083 M.
04

Calculate Ka

Using the equilibrium concentrations of C₆H₅CH₂COOH, C₆H₅CH₂COO⁻, and H₃O⁺, we can now determine the Ka value: Ka = [C₆H₅CH₂COO⁻][H₃O⁺] / [C₆H₅CH₂COOH] Substitute the equilibrium concentrations into the equation: Ka ≈ (2.09 × 10⁻³ M)(2.09 × 10⁻³ M) / 0.083 M Now, we can calculate Ka for phenylacetic acid: Ka ≈ 5.28 × 10⁻⁵

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Most popular questions from this chapter

Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base: $$ \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ A \(0.035 \mathrm{M}\) solution of ephedrine has a \(\mathrm{pH}\) of 11.33 . (a) What are the equilibrium concentrations of $\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}, \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+},\( and \)\mathrm{OH}^{-} ?$ (b) Calculate \(K_{b}\) for ephedrine.

Calculate \(\left[\mathrm{OH}^{-}\right]\) for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: (a) $\left[\mathrm{H}^{+}\right]=0.00010 \mathrm{M} ;(\mathbf{b})\left[\mathrm{H}^{+}\right]=7.3 \times 10^{-14} \mathrm{M} ;(\mathbf{c})\( a solution in which \)\left[\mathrm{OH}^{-}\right]$ is 100 times greater than \(\left[\mathrm{H}^{+}\right]\).

The average \(\mathrm{pH}\) of normal arterial blood is 7.40 . At normal body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{w}=2.4 \times 10^{-14} .\) Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right]\), and \(\mathrm{pOH}\) for blood at this temperature.

The odor of fish is due primarily to amines, especially methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)\). Fish is often served with a wedge of lemon, which contains citric acid. The amine and the acid react forming a product with no odor, thereby making the less-than-fresh fish more appetizing. Using data from Appendix \(D\), calculate the equilibrium constant for the reaction of citric acid with methylamine, if only the first proton of the citric acid \(\left(K_{a 1}\right)\) is important in the neutralization reaction.

The active ingredient in aspirin is acetylsalicylic acid \(\left(\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4}\right),\) a monoprotic acid with \(K_{a}=3.3 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\) What is the \(\mathrm{pH}\) of a solution obtained by dissolving one regular aspirin tablet, containing \(100 \mathrm{mg}\) of acetylsalicylic acid, in $200 \mathrm{~mL}$ of water?
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